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Chapter 4: Problem 15

A truck leaves a town traveling at a constant speed of \(\frac{55 \mathrm{mi}}{1 \mathrm{hr}}\). After \(20 \mathrm{~min}\), a car follows the same route traveling at a constant speed of \(\frac{60 \mathrm{mi}}{1 \mathrm{hr}}\). Find the time in minutes when the car will catch up with the truck. Find the distance traveled by the truck and the distance traveled by the car.

Short Answer

Expert verified
Time to catch up: 220 minutes. Distance traveled by truck: 220 miles. Distance traveled by car: 220 miles.

Step by step solution

01

- Convert Truck's Lead Time to Hours

The truck has a 20-minute head start. Convert this time to hours: \[ 20 \text{ minutes} = \frac{20}{60} \text{ hours} = \frac{1}{3} \text{ hours} \]
02

- Calculate Distance Traveled by the Truck Before the Car Starts

Use the truck's speed to find the distance it traveled in the time before the car started: \[ \text{Distance} = \text{Speed} \times \text{Time} \ \text{Distance} = 55 \frac{\text{mi}}{\text{hr}} \times \frac{1}{3} \text{ hr} = \frac{55}{3} \text{ mi} \]
03

- Set Up the Equation to Find When the Car Catches Up

Let \( t \) be the time in hours it takes for the car to catch up after it starts. Distance is the same for both when the car catches up. Therefore: \[ 55 \text{ mi/hr} \times \frac{1}{3} \text{ hr} + 55 \text{ mi/hr} \times t = 60 \text{ mi/hr} \times t \] Simplify to find \( t \).
04

- Solve for Time

Rearrange and solve the equation from Step 3: \[ \frac{55}{3} + 55t = 60t \ \frac{55}{3} = 5t \ t = \frac{55}{3 \times 5} = \frac{55}{15} = \frac{11}{3} \text{ hours} \] Convert to minutes: \[ \frac{11}{3} \text{ hours} \times 60 \text{ min/hour} = 220 \text{ minutes} \]
05

- Find the Distance Traveled

Using the time in hours (\( \frac{11}{3} \)) and the speeds of the truck and car, calculate the distances: \[ \text{Distance Traveled by the Truck} = 55 \text{ mi/hr} \times \frac{11}{3} \text{ hr} = \frac{605}{3} \text{ mi} \ \text{Distance Traveled by the Car} = 60 \text{ mi/hr} \times \frac{11}{3} \text{ hr} = 220 \text{ mi} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

distance-time relationship
Understanding the relationship between distance and time is crucial in solving travel problems. The fundamental idea is that distance is the product of speed and time. For example, if you move at a speed of 55 mi/hr for \(\frac{1}{3}\) of an hour, the distance you cover is: \[ \text{Distance} = 55 \frac{\text{mi}}{\text{hr}} \times \frac{1}{3} \text{ hr} = \frac{55}{3} \text{ mi} \] This linear relationship is straightforward and forms the basis for more complicated problems.
speed-time-distance calculations
Speed, time, and distance are interconnected. Knowing any two can help you find the third. Here, we use the truck鈥檚 head start and subsequent speeds to determine when the car catches up. The key steps are:
  • Convert units consistently (e.g., 20 minutes to hours).
  • Calculate the initial distance covered using the formula: \[ \text{Distance} = \text{Speed} \times \text{Time} \]
  • Set up an equation with both speeds and solve for the unknown time.
These principles apply universally in travel problems, allowing for consistent and accurate results.
linear equations in algebra
Linear equations are equations of the first degree, meaning they solve to a value 't'. In this scenario, algebra helps determine the time it takes for the car to catch up with the truck. We represent each vehicle鈥檚 travel as an equation and set them equal because they cover the same distance. So: \[ 55 \frac{\text{mi}}{\text{hr}} \times \frac{1}{3} \text{ hr} + 55 \frac{\text{mi}}{\text{hr}} \times t= 60 \frac{\text{mi}}{\hr} \times t \] Solving this equation isolates to: \[ t= \frac{55}{3 \times 5} = \frac{55}{15} = \frac{11}{3} \text{ hours} \] Linear equations like this are versatile tools in algebra, solving real-world problems efficiently.

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Most popular questions from this chapter

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